FUNCTIONS OF A REAL VARIABLE

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1 FUNCTIONS OF A REAL VARIABLE TSOGTGEREL GANTUMUR Abstract. We review some of the important concepts of single variable calculus. The discussions are centred around building and establishing the main properties of the elementary functions such as x a, exp x, log x, sin x, and arctan x. We start with an axiomatic treatment of real numbers. If the reader is willing to assume the basic properties of real numbers, then they can skip Section 1 in its entirety, and simply skim through Section 2. Contents 1. Ordered fields 1 2. The real number continuum 5 3. Limits and continuity 9 4. Differentiation Applications of differentiation Higher order derivatives Uniform convergence Power series The exponential and logarithm functions The sine and cosine functions Other trigonometric functions Antidifferentiation 44 Appendix A. Sets and functions Ordered fields In this and the following sections we state (one version of) the real number axioms, and derive the most fundamental properties of real numbers from them. The axioms can be thought of as the minimal requirements any logical system that claims itself to be the real number system must satisfy. It is possible to construct such systems by using other systems such as the natural number system, but we will not consider those here. The real number axioms can be divided into three groups. The set of real numbers is denoted by R, and for any two real numbers a, b R, their sum a + b R and product a b R are well defined. In other words, we assume the existence of two binary operations. Then the first group of axioms requires that R be a field with respect to these operations. This basically means that the addition and multiplication satisfy commutativity, associativity, and distributivity laws, that the numbers 0 and 1 exist and are distinct, and finally, that subtraction and division (by any nonzero number) can be defined. The second group of axioms adds more structure to the field, and demands that R be an ordered field. This means that any two real numbers a, b R satisfy one (and only one) of a < b, a = b, and b < a, that the binary relation < is transitive, and that a < b is preserved under addition of any number and under multiplication by a positive number. In this section, Date: September 4,

2 2 TSOGTGEREL GANTUMUR we will consider some of the consequences of the first two groups of axioms. The third group, that is an axiom on completeness of R, will be considered in the next section. Axiom 1 (Ordered field). (a) The addition operation satisfies the following properties. (i) a, b R then a + b = b + a. (ii) a, b, c R then (a + b) + c = a + (b + c). (iii) There exists an element 0 R such that a + 0 = a for each a R. (iv) For any a R there exists x R such that x + a = 0. (b) The multiplication operation satisfies the following properties. (i) a, b R then a b = b a. (ii) a, b, c R then (a b) c = a (b c). (iii) a, b, c R then a (b + c) = a b + a c. (iv) There exists an element 1 R with 1 0 such that a 1 = a for each a R. (v) For any a R not equal to 0, there exists x R such that x a = 1. (c) The binary relation < satisfies the following properties. (i) a, b R then one and only one of the following is true: a < b, a = b, or b < a. (ii) If a, b, c R satisfy a < b and b < c then a < c. (iii) If a, b, c R and a < b then a + c < b + c. (iv) If a, b R satisfy a > 0 and b > 0 then a b > 0. Remark 1.1. The relation a > b is defined as b < a. Similarly, a b means a < b or a = b, and a b means a > b or a = b. The product a b can be written simply as ab. Definition 1.2. Given a, b, c R with c 0, we define the difference a b R and the quotient a c R as the solutions to the equations b + x = a and cx = a, respectively. Then b) and f) of the following theorem guarantee that these concepts are well defined. We also define the opposite number (or negation) a = 0 a and the reciprocal (or inverse) a 1 = 1 a. Theorem 1.3 (Algebraic properties). For a, b, c R, we have the following. a) a + b = a implies b = 0 (uniqueness of 0). b) a + b = a + c implies b = c (subtraction of a). c) 0 a = 0. d) ab = 0 implies a = 0 or b = 0. e) ab = a and a 0 imply b = 1 (uniqueness of 1). f) ab = ac and a 0 imply b = c (division by a). Proof. a) By Axiom (a)(iii), we have b = 0 + b. Now let x be such that x + a = 0. Then the assumed property of b implies that b = 0 + b = (x + a) + b = x + (a + b) = x + a = 0. (1) b) Let x be such that x + a = 0. Then we have b = 0 + b = (x + a) + b = x + (a + b) = x + (a + c) = (x + a) + c = 0 + c = c. (2) c) Using the distributivity axiom, we observe that a = a 1 = a (0 + 1) = a 0 + a 1 = a 0 + a, (3) and part a) of the current theorem (i.e., uniqueness of 0) finishes the job. d) Suppose that a 0, and let x be such that xa = 1. Then we infer b = 1 b = (xa) b = x (ab) = x 0 = 0, (4) where in the last step we have used part c) of the current theorem. Exercise 1.4. Prove e) and f) of the preceding theorem. Exercise 1.5. Prove the following.

3 FUNCTIONS OF A REAL VARIABLE 3 (a) a b = a + ( b) for a, b R. (b) (ab) = ( a) b for a, b R. In particular, a = ( 1) a and ( a) ( a) = a a. (c) (ab) 1 = a 1 b 1 for a, b R \ {0}. Theorem 1.6 (Order properties). For a, b, c, d R, we have the following. a) a > 0 is equivalent to a < 0. b) a < b is equivalent to b a > 0. c) b < c and a > 0 imply ab < ac. d) b < c and a < 0 imply ab > ac. e) a 0 implies a a > 0. f) 0 < a < b and ac = bd > 0 imply 0 < d < c. Proof. a) If a > 0, then Axiom (c)(iii) yields 0 = a + ( a) > 0 + ( a) = a. Similarly, if a < 0, then 0 = a + a < 0 + a = a. b) By Axiom (c)(iii), a < b implies 0 = a + ( a) < b + ( a) = b a. Similarly, in the other direction, b a > 0 implies b = b a + a > 0 + a = a. c) By the preceding paragraph, b < c is the same as c b > 0, and then Axiom (c)(iv) implies that ac ab = a(c b) > 0, or ac > ab. f) Suppose that c < 0. Then c > 0, and hence (ac) = a ( c) > 0, or ac < 0. Since c = 0 would imply that ac = 0, we conclude that c > 0, and similarly, that d > 0. Now assume d c. Then ac < bc bd, contradicting ac = bd. Hence d < c. Exercise 1.7. Prove d) and e) of the preceding theorem. Definition 1.8. We introduce the following notations. (a, ) = {x R : x > a}, (, b) = {x R : x < b}. [a, ) = {x R : x a}, (, b] = {x R : x b}. (a, b) = (a, ) (, b). [a, b] = [a, ) (, b]. (a, b] = (a, ) (, b], [a, b) = [a, ) (, b). Exercise 1.9. Show that R \ (a, b) = (, a] [b, ) and R \ [a, b] = (, a) (b, ). Definition For a R, we define its modulus (or absolute value) by { a, for a 0, a = a, for a < 0. Remark One can think of a as the distance between the points a and 0 on the real line. The distance between two points a and b is defined to be a b. Exercise Prove the following. (a) a 0 for any a R, and a = 0 if and only if a = 0. (b) a 1 = 1 a for a 0. (c) ab = a b for a, b R. (d) a + b a + b for a, b R. (e) a b a b for a, b R. Having established the basic algebraic properties of R, we now turn to identifying some important subsets of R, such as the integers and the rational numbers. We have 0 R and 1 R with 0 1, thus obviously 1 0. Moreover, 1 < 0 would imply 1 > 0, and hence 1 = ( 1) 1 < ( 1) 0 = 0 by Theorem 1.6(c). Since 1 > 0 and 1 < 0 cannot be simultaneously true, we conclude that 0 < 1. Now, the number 2 = is distinct from each of 1, 0, and 1, because > = 1. We continue this process as 3 = 2 + 1, 4 = 3 + 1, (5)

4 4 TSOGTGEREL GANTUMUR..., constructing larger and larger numbers. The numbers obtained are of course the natural numbers. To make it more precise, we define an inductive subset of R as a subset A R with the property that 1 A and that x A implies x + 1 A. Then we define the set of natural numbers N as the intersection of all inductive subsets of R. Since inductivity is preserved under intersections (Exercise!), N is the smallest inductive subset of R. This property is the basis of proof by induction, and of definition by recurrence. We illustrate these by a couple of examples. Let n N, and let x k R for each k N with k n. In other words, x : N n R is a function with N n = {k N : k n} and x(k) = x k. Such a function is called a finite sequence of real numbers, an n-tuple, or an n-vector, and denoted by (x 1, x 2,..., x n ). Sets of the form {x k : k N n } are called finite sets. For an arbitrary set B R, a number m B is called a maximum of B if b m for all b B. If B admits a maximum, the maximum must be unique, because the existence of two maxima m, m B implies that m m and m m. Lemma Any nonempty finite set of real numbers admits a maximum. Proof. Let A N be the set of n N with the property that any set of the form {x k : k N n } admits a maximum. Clearly, 1 A since given any set {x 1 }, we can check that m = x 1 is a maximum of {x 1 }. Suppose that n A, that is, suppose that any set of the form {x k : k N n } admits a maximum. Let B = {x 1, x 2,..., x n+1 } be given. Then {x 1, x 2,..., x n } admits a maximum, which we denote by x i. Now if x i > x n+1, we set m = x i, and if x i x n+1, we set m = x n+1. Since m x n+1 and m x i x k for any k n, we see that m is the maximum of B. All this shows that 1 A, and that n A implies n + 1 A. Thus A is an inductive set, meaning that N A. Exercise Introduce the concept of minimum, and show that any nonempty set A N admits a minimum. Definition For a R and n N, the n-th power of a is the real number a n R defined by the recurrent formula { a n a, for n = 1, = a a n 1 (6), for n > 1. Fix a R, and let A R be the set of x R for which the power a x is uniquely defined by the preceding definition. Since the definition restricts itself to x N, we have A N. It is clear that 1 A. Moreover, if x A then x + 1 A, because x + 1 N and a x+1 is defined as a a x. Hence A is inductive, meaning that N A. Exercise (a) Informally, the factorials are defined by 1! = 1, 2! = 1 2, 3! = 1 2 3, etc. Give a definition of n! by using a recurrent formula. (b) Given x 1, x 2,..., x n, informally, we have x i = x 1, x i = x 1 + x 2, x i = x 1 + x 2 + x 3, etc. (7) Define i=1 i=1 n x i by using a recurrent formula. i=1 (c) Prove the binomial formula (a + b) n = n i=0 i=1 ( ) n a i b n i, (8) i where a, b R, n N, and ( ) n i = n! i!(n i)!, with the conventions x0 = 1 and 0! = 1.

5 (d) Prove the formula for a, b R and n N. FUNCTIONS OF A REAL VARIABLE 5 n 1 a n b n = (a b) a i b n 1 i, (9) The natural numbers are closed under addition and multiplication, but the equations a+x = b and ax = b are not always solvable. By adjoining the negative integers and 0 to N, the equation a + x = b can be solved. We let N 0 = {0} N, and define the set of integers by Z = N 0 { n : n N}. Then Z is closed under addition, subtraction, and multiplication, i.e., Z is a ring. We extend the power function to integers as follows. a n, for n N, a n = 1, for n = 0, (10) 1, for n < 0, a n where a R and n Z. Note that negative powers are not defined for a = 0. Exercise Prove the following. (a) a n a m = a n+m, (a n ) m = a nm, (ab) n = a n b n. (b) a n = a n (c) If 0 < a < b then a n < b n for n > 0 and a n > b n for n < 0. (d) If m < n then a m < a n for a > 1 and a m > a n for 0 < a < 1. Exercise Let A Z be bounded above, in the sense that there is b Z such that a < b for all a A. Show that A admits a maximum. Going further, the rational numbers are Q = { m n : m Z, n N}, which is closed under addition, subtraction, multiplication, and division by nonzero numbers. It is almost immediate that Q satisfies the ordered field axioms (Axiom 1) considered in this section. Therefore Axiom 1 is not stringent enough to differentiate R from Q. What we need is an additional axiom, that will be considered in the next section. i=0 2. The real number continuum The following notions will be used in the statement of the anticipated axiom. Definition 2.1. Given a set A R, a number s R is called the least upper bound or the supremum of A, and written as sup A = s, (11) if a s for all a A, and for any c < s there is a A with a > c. Example 2.2. For A = (0, 1] and B = (0, 1), we have sup A = sup B = 1. The set (1, ) does not have a supremum. Definition 2.3. A subset A R is called bounded above if there exists a number b R such that a < b for all a A. Example 2.4. The set {x x 2 : x R} is bounded above, while {n 2 : n N} is not. Now we are ready to state the only remaining axiom for real numbers. Axiom 2 (Continuum property, the least upper bound property). If A R is nonempty and bounded above, then there exists s R such that s = sup A.

6 6 TSOGTGEREL GANTUMUR Remark 2.5. Axiom 1 and Axiom 2 together pinpoint the real numbers completely, in the sense that if R and R both satisfy the aforementioned axioms, then there exists an order preserving field isomorphism between R and R. The basic reason behind this result is that each of R and R contains a copy of Q, and these two copies of Q can be naturally identified. Furthermore, the only field isomorphism from R onto itself is the identity map. Remark 2.6. Foreshadowing the detailed study that will occupy the rest of this section, here we want to include an informal discussion on the continuum property. In a certain sense, the continuum property combines two important characteristics of the real numbers. Firstly, it states that no single real number is larger than all the natural numbers, or equivalently, that there is no real number between 0 and all positive rational numbers. Loosely speaking, how large a real number can be is comparable to how large a natural number can be. This property is called the Archimedean property of the real numbers. The second characteristic of the real numbers that is embedded in the continuum property is basically the requirement that the real numbers have no gaps between them. This property is called the Cauchy completeness or the Cauchy property. Recall that by the ordered field axioms, R must contain Q as a subset. Then, the Cauchy property asserts that real numbers are used to fill in the gaps in Q, and the Archimedean property states that this filling in process is done efficiently, without adding any unnecessary elements. Theorem 2.7 (Archimedean property). a) Given any x R, there is n N such that x < n. b) Moreover, given any a R and b R with a < b, there exists q Q such that a < q < b. Proof. a) Suppose that there is x R satisfying n < x for all n N. Then by the least upper bound property, the supremum of N exists, i.e., s = sup N R. Now, by definition of supremum, there is m N such that m > s 1 2. This leads to contradiction, since m + 1 N and m + 1 > s b) Let δ = b a, and let n N be such that n > 1 δ. The set G = {k Z : k < bn} is by construction bounded above, and hence m = max G exists (Exercise 1.18). Let q = m n. Since m G, we have m < bn, meaning that q < b. Anticipating a contradiction, suppose that q a. This would mean that m an, or m + 1 an + 1 = (a + 1 n )n < (a + δ)n = bn. Therefore, we have m + 1 G, which contradicts the maximality of m. In the remainder of this section, we explore further consequences of the continuum property. We start by introducing the notion of infinite sequences. Definition 2.8. A real number sequence is a function x : N R, which is usually written as {x n } = {x 1, x 2,...}, with x n = x(n). We say that a sequence {x n } converges to x R, if for any given ε > 0, there exists an index N such that If {x n } converges to x, we write x n x ε for all n N. (12) lim x n = x, or lim x n = x, or x n x as n. (13) n In some contexts, the sequence {x n } is identified with the set {x n : n N}. For instance, {x n } Ω with some Ω R means that x n Ω for all n. Moreover, often times one considers sequences such as {a 0, a 1,...}, whose indices start with n = 0. Definition 2.9. If a sequence does not converge to any number, then the sequence is said to diverge. A special type of divergence occurs when the divergence is caused by growth, rather than oscillation. More precisely, we say that {a n } diverges to, and write lim a n =, or lim a n =, or a n as n, (14) n

7 FUNCTIONS OF A REAL VARIABLE 7 if for any given number M, the sequence is eventually larger than M, that is, only finitely many terms of {a n } stay in (, M). Divergence to can be defined in an obvious manner. Example (a) A constant sequence is a sequence whose terms are all equal to each other. An example is the sequence {1, 1, 1,...} whose n-th term is a n = 1. (b) An arithmetic progression is a sequence whose n-th term satisfies the formula a n = kn+b, where k and b are constants. For example, {5, 8, 11, 14,...} is an arithmetic progression, with k = 3 and b = 2. Note that we have chosen b = 2 so that the first term 5 corresponds to n = 1 in the formula a n = kn + b. However, this is not necessary. We could have taken b = 5, and specified that we start the sequence at the index n = 0. (c) Similarly, a geometric progression is a sequence with the general term a n = bq n, where q and b are constants. For example, {1, 1 2, 1 4, 1 8,...} is a geometric progression, with q = 1 2 and b = 1, and with the understanding that the first term of the sequence corresponds to n = 0 in the formula a n = bq n. (d) The sequence {1, 4, 9, 16,...}, with a n = n 2, is called the sequence of square numbers. (e) The Fibonacci sequence is {0, 1, 1, 2, 3, 5, 8, 13, 21,...}, where each term (except the first two) is the sum of the two terms immediately preceding it. In other words, we have the recurrent formula a n = a n 1 + a n 2 for n 3. Example (a) We want to show that if x > 1 then lim x n =. To this end, we start with the binomial formula (a + b) n = a n + na n 1 b nab n 1 + b n, (15) and substitute a = 1 and b = x 1 > 0, which gives x n = (1 + b) n = 1 + nb nb n 1 + b n 1 + nb, (16) since all the terms of the sum are positive. Now, given a large number M, choose N so that 1 + Nb > M. For instance, N > M/b would be sufficient. Then for all n > N, we would have x n 1 + nb > 1 + Nb > M. (17) This means, by definition, that lim x n =. (b) Let us show that if 0 < x < 1 then lim x n = 0. So let ε > 0 be given. We know that lim y n =, where y = 1 x. By definition, for any given M, there exists an index N such that y n > M for all n > N. Let N be such an index that corresponds to the choice M = 1 ε. Then, for all n > N, we have xn = x n = 1 y n < 1 M = ε. This shows that lim x n = 0. Remark The limit of a sequence depends only on behaviour at n =, in the sense that if lim a n = a, then after an arbitrary modification (or removal) of finitely many terms of {a n }, we would still have lim a n = a. To change the limit behaviour one would have to modify infinitely many terms. Exercise Prove the following. (a) If 1 < x < 1 then lim x n = 0. (b) If x > 1 then lim xn n =. (c) If x > 1 and a N then lim xn n = and lim n a x n = 0. a (d) lim xn n! = 0 for any x R. Exercise Prove the following. (a) lim a n = a if and only if lim a n a = 0. (b) If lim a n = and b n a n for all n then lim b n =. (c) If {a n } converges then it is bounded, i.e., there is M R such that a n M for all n.

8 8 TSOGTGEREL GANTUMUR Theorem 2.15 (Monotone convergence). Let {x n } R be a sequence that is nondecreasing and bounded above, in the sense that x n x n+1 M for each n, (18) and with some constant M R. Then there is x M such that x n x as n. Proof. Let x = sup{x n }, and let ε > 0. Then there is N such that x ε < x N. Since {x n } is nondecreasing, we have x ε < x n x for all n N. This means that {x n } converges to x. The inequality x M is obvious because x is the least upper bound of {x n }. Exercise Show that every nonincreasing, bounded below sequence converges. Theorem 2.17 (Nested intervals principle). Let {a n } and {b n } be two sequences such that [a 0, b 0 ] [a 1, b 1 ]... [a n, b n ]..., (19) with a n < b n for each n N. Then the intersection n N [a n, b n ] is nonempty. In addition, if b n a n 0 as n, then n N [a n, b n ] consists of a single point. Proof. We have a 0 a 1... a n < b n... b 1 b 0, (20) which makes it clear that {a n } is nondecreasing and {b n } is nonincreasing. Since both of these sequences are bounded, by the monotone convergence theorem (Theorem 2.15), there exist a and b such that a m a and b m b as m. Given any m, we have a m a n < b n b m whenever n m. This implies that a and b are both in the interval [a m, b m ] for any m. In addition, if b n a n 0, we must have a = b. Theorem 2.18 (Bolzano-Weierstrass). Let {x n } [a, b]. {x nk } {x n } that converges to some point x [a, b]. Then there is a subsequence Proof. Let us subdivide the interval [a, b] into two subintervals [a, a+b a+b 2 ] and [ 2, b]. Then at least one of these subintervals must contain infinitely many terms from the sequence {x n }. Pick one such subinterval, and call it [a 1, b 1 ]. Obviously, we have b 1 a 1 = b a 2. Now we subdivide [a 1, b 1 ] into two halves [a 1, a 1+b 1 2 ] and [ a 1+b 1 2, b 1 ], one of which must contain infinitely many terms from {x n }. Recall that interval [a 2, b 2 ]. Of course, we have b 2 a 2 = b a 4. We continue this process indefinitely, and obtain a sequence of intervals [a, b] [a 1, b 1 ]... [a m, b m ]..., (21) with each [a m, b m ] containing infinitely many terms from the sequence {x n }, and satisfying b m a m = 2 m (b a). Now the nested intervals principle (Theorem 2.17) implies that there exists a R such that a [a n, b n ] for all n N. Hence a a n 2 n (b a) for all n. Let n 0 = 1, and for k N, let n k be an index such that n k > n k 1 and that x nk [a k, b k ]. Such n k exists since [a k, b k ] contains infinitely many terms from {x n }. Then we have x nk a x nk a k + a k a 2 k (b a) + 2 k (b a), (22) which shows that the sequence {x nk } converges to a. Theorem 2.19 (Cauchy s criterion). Let {x n } R be a Cauchy sequence, in the sense that Then {x n } is convergent. x n x m 0, as min{n, m}. (23) Proof. Let N be such that x n x N 1 for all n N. Then we have and therefore x n x N + 1 for all n N, (24) x n max{ x 1,..., x N 1, x N + 1} for all n, (25)

9 FUNCTIONS OF A REAL VARIABLE 9 meaning that {x n } is bounded. By the Bolzano-Weierstrass theorem (Theorem 2.18), there is a subsequence {x nk } {x n } that converges to some point x R. So far we only have shown that a subsequence of {x n } converges to x. Now we will show that the whole sequence {x n } indeed converges to x. To this end, let ε > 0, and let N be such that x n x m ε for all n N and m N. Moreover, let k N be large enough that x nk x ε. Then for m N, we have x m x x m x nk + x nk x 2ε, (26) which shows that the entire sequence {x n } converges to x. 3. Limits and continuity In this section, we will study continuous functions. Intuitively, a continuous function f sends nearby points to nearby points, i.e, if x is close to y then f(x) is close to f(y). This intuition can be made precise by saying that if a sequence x k converges to x, then f(x k ) converges to f(x). Before doing that, we prove a preliminary result to handle limits. Recall from Definition 2.8 that a sequence {x n } converges to x R, if for any given ε > 0, there exists an index N such that x n x ε for all n N. (27) If {x n } converges to x, we write lim n x n = x or x n x as n. Theorem 3.1. Let a n a and b n b as n. Then the following are true. a) a n ± b n a ± b as n. b) a n b n ab as n. c) If a 0, then a n = 0 for only finitely many indices n, and after the removal of those zero terms from the sequence {a n }, we have lim 1 a n = 1 a. d) a n b n implies a b. e) If {x n } is a sequence satisfying a n x n b n for all n, and if a = b, then lim x n = a. Proof. b) We have a n b n ab = a n (b n b) + (a n a)b, and so a n b n ab a n b n b + a n a b for all n. (28) By choosing n large enough, we can make b n b and a n a b as small as we want. The question is if can do the same for the product a n b n b. We claim that {a n } is bounded, i.e., there is M R such that a n M for all n. Indeed, since {a n } converges to a, taking ε = 1 in the definition of convergence, there exists an index N such that a n a 1 for all n N. Hence a n = a n a + a a n a + a 1 + a for all n N, and if we take M = max{ a + 1, a 1, a 2,..., a N 1 }, then a n M for all n. Now (28) yields a n b n ab M b n b + b a n a for all n. (29) Let ε > 0 be given. Let N be such that M b n b ε 2 for all n N, and let N be such that b a n a ε 2 for all n N. This is possible since b n b and a n a as n. Now we set N = max{n, N }. Then we have a n b n ab M b n b + b a n a ε 2 + ε 2 = ε for all n N, (30) which means that a n b n ab as n. e) Let ε > 0 be given. Let N be such that a n a ε for all n N, and let N be such that b n b ε for all n N. We set N = max{n, N }. Then we have and x n a = x n b b n b b n b ε for all n N, (31) a x n a a n a n a ε for all n N, (32)

10 10 TSOGTGEREL GANTUMUR which imply x n a = max{x n a, a x n } ε for all n N. (33) By definition, this means that x n a as n. Exercise 3.2. Prove a), c) and d) of the preceding theorem. Example 3.3. Let us try to compute the limit of 3n+1 2n+5 as n. If we approach naively, we get 3n + 1 lim n 2n + 5 = lim 1 (3n + 1) lim = 0, (34) n n 2n + 5 which is nonsense. The error is in the first step, where we attempt to apply Theorem 3.1b). This is not justified, because lim(3n + 1) does not exist. A correct way to proceed is to write 3n + 1 2n + 5 = n n and to note that lim(3 + 1 n ) = 3 and lim(2 + 5 n ) = 2 by Theorem 3.1a). Hence lim 1 = by Theorem 3.1c). n Therefore lim(3 + 1 n ) n This process is usually written as = = 3 2 lim 3n + 1 2n + 5 = lim n n = (3 + 1 n ) , (35) n by Theorem 3.1b). = 3 + lim 1 n 2 + lim 5 n = 3 2. (36) Exercise 3.4. Prove the following. (a) If lim a n = and lim bn a n = 0, then lim(a n ± b n ) =. (b) lim a n = if and only if lim 1 a n = 0 and {a n } is eventually positive. (c) If lim a n = 0 and {b n } is bounded, then lim(a n b n ) = 0. Recall that a sequence {a n } is bounded if there exists a number M such that a n M for all n. We define continuous functions as the ones that send convergent sequences to convergent sequences. This is sometimes called the sequential criterion of continuity. Definition 3.5. Let K R be a set. A function f : K R is called continuous at x K if f(x n ) f(x) as n for every sequence {x n } K converging to x. Example 3.6. (a) Let c R, and let f : R R be the function given by f(y) = c for y R. Then f is continuous at every point x R, since for any sequence {x n } R converging to x, we have f(x n ) = c c = f(x) as n. (b) Let f : R R be the function given by f(y) = y for y R. Then f is continuous at every point x R, because given any sequence {x n } R converging to x, we have f(x n ) = x n x = f(x) as n. We now confirm an intuitive property of continuous functions, namely that if f is continuous at x then for all points y close to x the value f(y) is close to f(x). Lemma 3.7. Let K R be a set. Then f : K R is continuous at x K if and only if for any ε > 0 there exists δ > 0 such that y (x δ, x+δ) K implies f(x) ε < f(y) < f(x)+ε. Proof. Let f be continuous at x K, and let ε > 0. Suppose that no such δ exists, i.e., that there is a sequence {x n } K converging to x, with f(x n ) f(x) ε for all n. Since f is continuous at x, we have f(x n ) f(x) as n. In particular, there is an index N such that f(x) f(x N ) < ε, which is a contradiction.

11 FUNCTIONS OF A REAL VARIABLE 11 In the other direction, assume that x K and that for any ε > 0 there exists δ > 0 such that y (x δ, x + δ) K implies f(x) ε < f(y) < f(x) + ε. In particular, there is a positive sequence {δ n } such that y (x δ n, x + δ n ) K implies f(x) 1 n < f(y) < f(x) + 1 n. Let {x m } K be a sequence converging to x. Then we can choose a sequence of indices m 1, m 2,..., such that x m (x δ n, x+δ n ) for all m m n, that is, f(x) 1 n < f(x m) < f(x)+ 1 n for all m m n. This shows that f(x m ) f(x) as m. The following result shows that continuity is a local property. Lemma 3.8. Let f : K R with K R, and let g = f (a,b) K for some (a, b). Then f is continuous at x (a, b) K if and only if g is continuous at x. Proof. Suppose that f is continuous at x (a, b) K. Then by definition, f(x n ) f(x) as n for every sequence {x n } K converging to x. In particular, this is true for every sequence {x n } (a, b) K converging to x. Since g = f on (a, b) K, g is continuous at x. Now suppose that g is continuous at x (a, b) K, i.e., that f(x n ) = g(x n ) g(x) = f(x) as n for every sequence {x n } (a, b) K converging to x. Let {x n } K be a sequence converging to x. Then there exists N such that x n (a, b) K for all n N, and hence the sequence {f(x N ), f(x N+1 ),...} converges to f(x). This is the same as saying that the full sequence {f(x n )} converges to f(x), meaning that f is continuous at x. Our next step is to combine known continuous functions to create new continuous functions. Definition 3.9. Given two functions f, g : Ω R, with Ω R, we define their sum, difference, product, and quotient by (f ± g)(x) = f(x) ± g(x), (fg)(x) = f(x)g(x), and (f ) f(x) (x) = g g(x), (37) for x Ω, where for the quotient definition we assume that g does not vanish anywhere in Ω. Furthermore, we define the function f by f (z) = f(z), for x Ω. (38) Theorem Let Ω R, and let f, g : Ω R be functions continuous at x Ω. Then the following are true. a) The sum and difference f ± g, the product fg, and the modulus f are all continuous at x. b) The function 1 f is continuous at x, provided that f(x) 0. c) Suppose that U R is a set satisfying g(ω) U, the latter meaning that y Ω implies g(y) U. Let F : U R be a function continuous at g(x). Then the composition F g : Ω R, defined by (F g)(y) = F (g(y)), is continuous at x. Proof. The results are immediate from the definition of continuity. For instance, let us prove that fg is continuous at x. Thus let {x n } Ω be an arbitrary sequence converging to x. Then f(x n ) f(x) and g(x n ) g(x) as n, and Theorem 3.1 gives f(x n )g(x n ) f(x)g(x) as n. Hence fg is continuous at x. Exercise Prove b) and c) of the preceding theorem. Definition A function f : Ω R is called continuous in Ω, if f is continuous at each point of Ω. The set of all continuous functions in Ω is denoted by C (Ω). Exercise Show that if f, g C (Ω), then f ± g, fg, f C (Ω). Example (a) Recall from Example 3.6 that the constant function f(x) = c (where c R) and the identity map f(x) = x are continuous in R. Then by Theorem 3.10a),

12 12 TSOGTGEREL GANTUMUR any monomial f(x) = ax n with a constant a R, is continuous in R, since we can write ax n = a x x. Applying Theorem 3.10a) again, we conclude that any polynomial p(x) = a 0 + a 1 x a n x n, (39) where a 0,... a n R are the coefficients, as a function p : R R, is continuous in R. (b) Let p and q be polynomials, and let Z = {x R : q(x) = 0} be the set of real roots of q. Then by Theorem 3.10b), the function r : R \ Z R given by r(x) = p(x) q(x) is continuous in R \ Z. The functions of this form are called rational functions. For instance, f(x) = x2 +1 x 1 is a rational function defined for x R \ {1}. Exercise Show that functions of the form r 1(x)+r 2 ( x ) r 3 (x)+r 4 ( x ) are continuous in an appropriate subset of R, where r 1, r 2, r 3, and r 4 are all rational functions. The following theorem was proved by Bernard Bolzano in Theorem 3.16 (Intermediate value). Let a, b R satisfy a < b, and let f : [a, b] R be continuous in [a, b]. Then for any y R satisfying min{f(a), f(b)} y max{f(a), f(b)}, there exists x [a, b] such that f(x) = y. Proof. We apply what is known as the bisection method. We define two sequences {a n } and {b n } as follows. First, set a 0 = a and b 0 = b. Obviously, the value y lies between f(a 0 ) and f(b 0 ), that is, y [f(a 0 ), f(b 0 )] [f(b 0 ), f(a 0 )]. Let c 0 = 1 2 (a 0 + b 0 ). Then at least one of y [f(a 0 ), f(c 0 )] [f(c 0 ), f(a 0 )] and y [f(c 0 ), f(b 0 )] [f(b 0 ), f(c 0 )] must hold. If the former holds, we set a 1 = a 0 and b 1 = c 0. Otherwise, we set a 1 = c 0 and b 1 = b 0. In any case, we have y [f(a 1 ), f(b 1 )] [f(b 1 ), f(a 1 )], and b 1 a 1 = 1 2 (b 0 a 0 ). By repeating this process, we get {a n } and {b n } such that [a 0, b 0 ] [a 1, b 1 ]... [a n, b n ]..., (40) with y [f(a n ), f(b n )] [f(b n ), f(a n )] and b n a n = 2 n (b a) for each n N. Then by the nested intervals principle (Theorem 2.17), n N [a n, b n ] is nonempty and consists of a single point. Let us denote this point by x. Since a n x and b n x as n, by continuity of f we have f(a n ) f(x) and f(b n ) f(x) as n. In particular, for any given ε > 0, there exists N such that f(a n ) f(x) ε and f(b n ) f(x) ε for all n N. This means that f(b n ) f(a n ) 2ε for all n N, and hence y f(a n ) 2ε for all n N. Finally, by the triangle inequality we have f(x) y f(x) f(a N ) + f(a N ) y 3ε. In other words, for any ε > 0 we have f(x) y 3ε. This shows that f(x) = y, since f(x) y would imply that f(x) y > 0. Example Given y 0, let us try to solve the equation x 2 = y. Consider the function f(x) = x 2 in the interval [0, b], where b = max{1, y}. Obviously, f is continuous in [0, b], and since y max{1, y 2 }, we have f(0) y f(y). By the intermediate value theorem (Theorem 3.16), there exists x [0, b] such that f(x) = y, that is, the equation x 2 = y has a solution in the interval [0, b]. This solution is in fact the unique solution of x 2 = y in [0, ). Indeed, 0 x < z implies x 2 < z 2, and hence x 2 = z 2 with x, z 0 is possible only when x = z. The function y x is called the arithmetic (or principal) square root function, and denoted by x = y. It is clear that the arithmetic square root function is the inverse of f : [0, ) [0, ) given by f(x) = x 2. More generally, let F : R [0, ) be given by F (x) = x 2. Then we have F [0, ) = f, and F (x) = f( x ) for all x R. This means that x 2 = z 2 implies x = z, and hence x 2 = z 2 if and only if x = z or x = z. To conclude, the equation x 2 = y has no solution for y < 0, exactly one solution for y = 0, and exactly two solutions x = y and x = y for y > 0. Exercise Investigate the equation x n = y where n N.

13 FUNCTIONS OF A REAL VARIABLE 13 Corollary Let f C ([a, b]) be a strictly increasing function in the sense that x < y implies f(x) < f(y). Then f is injective, the image of f is f([a, b]) = [f(a), f(b)], and the inverse f 1 : f([a, b]) [a, b] is continuous and strictly increasing. Proof. Considered as a function f : [a, b] f([a, b]), f is obviously surjective. Since x y implies f(x) f(y), f is injective. Hence there exists the inverse f 1 : f([a, b]) [a, b]. Moreover, x y implies f(x) f(y), and so by contrapositive, f 1 is strictly increasing. We have [f(a), f(b)] f([a, b]), since by the intermediate value theorem (Theorem 3.16), the equation f(x) = y has a solution for every y [f(a), f(b)]. On the other hand, a x b implies f(a) f(x) f(b), meaning that f([a, b]) = [f(a), f(b)]. It remains to show that f 1 is continuous. To this end, let {y n } [f(a), f(b)] be such that y n y for some y [f(a), f(b)]. Let ε > 0 be given, and let x = f 1 (y). Then with α ε = f(max{a, x ε}) and β ε = f(min{b, x + ε}), we have f 1 (y n ) [x ε, x + ε] whenever y n [α ε, β ε ]. Note that α ε < y unless y = f(a), and β ε > y unless y = f(b). Thus there exists N such that y n [α ε, β ε ] for all n N, which shows that f 1 (y n ) x as n. Exercise Let f C ([a, b]) be a strictly decreasing function in the sense that x < y implies f(x) > f(y). Show that f is injective, and that the inverse f 1 : f([a, b]) [a, b] is continuous and strictly decreasing. Example (a) Fix n N, and consider the power function f(x) = x n. This is a continuous and strictly increasing function in the range [0, ). Given any b > 0, the restriction f [0,b] is in particular strictly increasing, and so its inverse g b : [0, b n ] [0, b] is also strictly increasing and continuous. Now if we consider g c with c > b, then g c must agree with g b on their common domain, that is, g c (y) = g b (y) for y [0, b n ], since g b (f(x)) = x and g c (f(x)) = x for x [0, b]. Therefore we can define the function g : [0, ) [0, ) by g(y) = g b (y) with b > max{1, y} for y [0, ), and this function satisfies g(f(x)) = x for x [0, ), i.e., g is the inverse of f [0, ). Moreover, g is continuous and strictly increasing. Of course, g is the arithmetic n-th root function, denoted by n y g(y). (b) With the help of the n-th root function, we also define the rational power as x m n = ( n x ) m for x 0, m Z, n N. (41) That this definition is unambiguous can be seen as follows. We have (( n x) m ) n = ( n x) mn = (( n x) n ) m = x m, and hence ( n x) m = n x m. By writing m n = mk mk nk with some k N, we get x nk = ( nk x) mk. We need to show that this is equal to ( n x) m. Since (( nk x) k ) n = ( nk x) nk = x, we have ( nk x) k = n x. Therefore we conclude that ( nk x) mk = (( nk x) k ) m = ( n x) m. As the composition of two continuous functions, the rational power function w(x) = x m n is a continuous function of x 0. Moreover, w is strictly increasing, provided m n > 0, since it is the composition of two strictly increasing functions. Another important property is for a, b Q with a < b, we have x > 1 = x a < x b and 0 < x < 1 = x a > x b. (42) Indeed, by writing a = m n and b = k n, the question is reduced to comparing the integer powers ( n x) m and ( n x) k. n (c) We claim that for any x > 0, x 1 as n. This result will be used later to define the power x a for any a R. Let x > 1. Then n x > 1 for all n. Suppose that there exists some ε > 0 such that n x > 1 + ε for all n. This would imply that x = (1 + ε) n 1 + εn for all n, which is impossible by the Archimedean property. Hence n x 1 as n, for x > 1. The case 0 < x < 1 is given as an exercise below.

14 14 TSOGTGEREL GANTUMUR Exercise Prove the following. (a) For odd n, the inverse of f(x) = x n can be defined on all of R. (b) The function f(x) = x q with rational q < 0 is a strictly decreasing function of x 0. (c) For x, y 0 and p, q Q, there hold that Exercise Prove the following. (a) If 0 < x < 1 then n x 1 as n. (b) n n 1 as n. (c) If a N then n n a 1 as n. x p x q = x p+q, (x p ) q = x pq, (xy) q = x q y q. (43) Remark Let s = 2. We claim that s Q, which would mean that Q is strictly contained in R. To show this, suppose that s Q, i.e., that s = m n with m N and n N. Thus we have ( m n )2 = 2 or m 2 = 2n 2. This implies that m is divisible by 2, i.e., m = 2m 1 for some m 1 N, which in turn yields n 2 = 2m 2 1. Hence n = 2n 1 for some n 1 N, and this leads to m 1 = 2m 2 for some m 2 N. We can repeat this process indefinitely, arriving at the conclusion that for any k N there exists a N such that m = 2 k a. However, since a 1, by choosing k large enough we can ensure that 2 k a 2 k > m, which is a contradiction. The numbers in R \ Q, such as 2, are called irrational numbers. The following fundamental theorem was established by Weierstrass in Theorem 3.25 (Extreme value). Let f C ([a, b]). Then there exists c [a, b] such that f(x) f(c) for all x [a, b]. Proof. For n N, let Q n = { k m : k Z, m N, m n} [a, b]. It is clear that Q n is a finite set, and therefore f takes its maximum over Q n, i.e., there exists x n Q n such that f(q) f(x n ) for all q Q n. Since {x n } [a, b], the Bolzano-Weierstrass theorem (Theorem 2.18) guarantees the existence of a subsequence {x nk } {x n } that converges to some point c [a, b]. The sequence {f(x n )} is nondecreasing, in the sense that f(x n ) f(x n+1 ) for all n. Thus f(c) f(x n ) for all n, and hence f(c) f(q) for all q Q [a, b]. We claim that f(x) f(c) for all x [a, b]. Suppose that there is x [a, b] with f(x) > f(c). Then by continuity there exists ε > 0 small enough, such that y (x ε, x + ε) implies f(y) > f(c). However, there exists q Q (x ε, x + ε), contradicting the fact that f(c) f(q) for all q Q [a, b]. 4. Differentiation Let us consider the function f(x) = x 2 when x is very close to some given point x R. Putting h = x x, which is assumed to be small, we can write that is, x 2 = (x + h) 2 = x 2 + 2x h + h 2, (44) f(x) f(x ) = x 2 x 2 = 2x h + h 2 = (2x + h)h. (45) Intuitively speaking, this means that when h small, f(x + h) f(x ) is basically equal to the linear function l(h) = 2x h. This leads to the concept of derivative. Given a function f : (a, b) R with a < b, and a point x (a, b), the idea is to require f(x + h) f(x ) = (λ + e(h))h, (46) where λ R is a constant, and e(h) is a function of h that can be made arbitrarily close to 0 by choosing h small enough. In other words, e is continuous at h = 0 with e(0) = 0. An example of such a function is e(h) = h as in (45). If (46) holds, then for h R with h small, f(x + h) f(x ) is equal to the linear function l(h) = λh, up to the error e(h)h. The

15 FUNCTIONS OF A REAL VARIABLE 15 following definition was introduced by Carathéodory in 1950, and is a refined version of the definition given by Weierstrass in Definition 4.1. A function f : (a, b) R is said to be differentiable at x (a, b), if there exists a function g : (a, b) R, which is continuous at x, such that f(y) = f(x) + g(y)(y x), y (a, b). (47) We call the value g(x) the derivative of f at x, and write f (x) df (x) := g(x). (48) dx If f is differentiable at each x K for some K (a, b), then we say that f is differentiable in K, and consider the derivative as a function f : K R sending x to f (x). It is immediate from (47) that if f is differentiable at x then f is continuous at x. The following lemma gives a sequential criterion of differentiability. This criterion was used as a definition by Cauchy in Lemma 4.2. A function f : (a, b) R is differentiable at x (a, b) if and only if there exists a number λ R such that f(x n ) f(x) λ as n, (49) x n x for every sequence {x n } (a, b) \ {x} converging to x. Proof. Let f be differentiable at x. Then by (47), we have g(y) = f(y) f(x) y x for y (a, b) \ {x}. (50) Since g is continuous at x, for any sequence {x n } (a, b) \ {x} converging to x, we have g(x n ) = f(x n) f(x) x n x λ := g(x) as n. (51) This establishes the only if part of the lemma. Now suppose that there exists λ R such that (49) holds for every sequence {x n } (a, b) \ {x} converging to x. Then we define a function g : (a, b) R by g(y) = f(y) f(x) y x for y (a, b) \ {x}, and g(x) = λ. (52) This function satisfies (47) by construction. It remains to show that g is continuous at x. Let {x n } (a, b) be a sequence converging to x. Suppose that we created a new sequence {x m} (a, b) \ {x} by removing every occurrence of x from {x n }. There are two possibilities. The first possibility is that {x m} is a finite sequence. In this case, there exists some N such that x n = x for all n N, and hence it is trivial to observe that g(x n ) g(x) as n. The second possibility is that {x m} is an infinite sequence. In this case, by (49) we have g(x m) g(x) as m, that is, for any ε > 0, there exists M such that g(x m) g(x) ε for all m M. Now if we let N to be the index of {x n } corresponding to the index M in {x m}, then it is clear that g(x n ) g(x) ε for all n N because for n N we have either x n = x m for some m M or x n = x. Example 4.3. (a) Let c R, and let f(x) = c be a constant function. Then since f(y) = f(x) + 0 (y x) for all x, y, we get f (x) = 0 for all x. (b) Let a, c R, and let f(x) = ax + c be a linear (also known as affine) function. Since f(y) = f(x) + a(y x) for all x, y, we get f (x) = a for all x.

16 16 TSOGTGEREL GANTUMUR (c) Let f(x) = 1 x, and for y R \ {0, x} define g(y) = 1 y 1 x y x = 1 xy. (53) As long as x 0, upon defining g(x) = 1, the function g(y) = 1 x 2 x 1 y becomes continuous at y = x, and therefore f is differentiable at x with f (x) = ( 1 ) 1 = x x 2 (x 0). (54) (d) Let us try to differentiate f(x) = x at x = 0. With x n = 1 n for n N, we have {x n } R \ {0} and x n 0 as n. On one hand, we get f(x n ) f(0) x n lim = lim = 1, (55) n x n 0 n x n but on the other hand, with y n = x n, we infer f(y n ) f(0) y n x n lim = lim = lim = 1. (56) n y n 0 n y n n x n The definition of derivative requires these two limits to be the same, and thus we conclude that f(x) = x is not differentiable at x = 0. (e) Consider the differentiability of f(x) = 3 x at x = 0. Let x n = 1. It is obvious that n 3 x n 0 and x n 0. We have f(x n ) f(0) x n 0 = 3 x n x n = n 2, (57) which diverges as n. Hence f(x) = 3 x is not differentiable at x = 0. Exercise 4.4. Show that f(x) = x n is differentiable in R, for n N, with f (x) = nx n 1. The following result shows that differentiability is a local property. Lemma 4.5. Let f : (a, b) R, and let g = f (c,d) for some (c, d) (a, b). Then f is differentiable at x (c, d) if and only of g is differentiable at x. Moreover, if f is differentiable at x (c, d), then f (x) = g (x). Proof. Suppose that f is differentiable at x (c, d). Then by definition, there is a function f : (a, b) R, continuous at x, with f (x) = f(x), such that Since g(y) = f(y) for y (c, d), we have f(y) = f(x) + f(y)(y x), y (a, b). (58) g(y) = g(x) + f(y)(y x), y (c, d). (59) This shows that g is differentiable at x with g (x) = f(x) = f (x). Now suppose that g is differentiable at x (c, d). Then there is a function g : (c, d) R, continuous at x, with g (x) = g(x), such that g(y) = g(x) + g(y)(y x), y (c, d). (60) If we extend g as { g(y), for y (c, d), f(y) = f(y) f(x) y x, for y (a, b) \ (c, d), then since f and g agree on (c, d), we have (61) f(y) = f(x) + f(y)(y x), y (a, b). (62)

17 FUNCTIONS OF A REAL VARIABLE 17 Since f = g on (c, d), by locality of continuity f is continuous at x. Hence f is differentiable at x with f (x) = f(x) = g(x) = g (x). We now investigate differentiability of various combinations of differentiable functions. Theorem 4.6. Let f, g : (a, b) R be functions differentiable at x (a, b). following are true. a) The sum and difference f ± g are differentiable at x, with These are called the sum and difference rules. b) The product fg is differentiable at x, with Then the (f ± g) (x) = f (x) ± g (x). (63) (fg) (x) = f (x)g(x) + f(x)g (x). (64) This is called the product rule. c) If F : (c, d) R is a function differentiable at g(x), with g((a, b)) (c, d), then the composition F g : (a, b) R is differentiable at x, with (F g) (x) = F (g(x))g (x). (65) This is called the chain rule. d) If f : (a, b) f((a, b)) is bijective and f (x) 0, then the inverse f 1 : f((a, b)) (a, b) is differentiable at y = f(x), with (f 1 ) (y) = 1 f (x). (66) Proof. b) By definition, there is a function f : (a, b) R, continuous at x, satisfying f(y) = f(x) + f(y)(y x), y (a, b), (67) and f (x) = f(x). Similarly, there is a function g : (a, b) R, continuous at x, and with g (x) = g(x), such that By multiplying (67) and (68), we get g(y) = g(x) + g(y)(y x), y (a, b). (68) f(y)g(y) = f(x)g(x) + g(x) f(y)(y x) + f(x) g(y)(y x) + f(y) g(y)(y x) 2 = f(x)g(x) + [g(x) f(y) + f(x) g(y) + f(y) g(y)(y x)](y x). The expression in the square brackets, as a function of y, is continuous at y = x, with [g(x) f(y) + f(x) g(y) + f(y) g(y)(y x)] y=x = g(x) f(x) + f(x) g(x) = g(x)f (x) + f(x)g (x), which shows that fg is differentiable at x, and that (64) holds. c) Since F is differentiable at g(x), by definition, there is a function F : (c, d) R, continuous at g(x), and with F (g(x)) = F (g(x)), such that Plugging z = g(y) into (71), we get (69) (70) F (z) = F (g(x)) + F (z)(z g(x)), z (c, d). (71) F (g(y)) = F (g(x)) + F (g(y))(g(y) g(x)) = F (g(x)) + F (g(y)) g(y)(y x), (72) where in the last step we have used (68). The function y F (g(y)) g(y) is continuous at y = x, with F (g(x)) g(x) = F (g(x))g (x), which confirms that F g is differentiable at x, and that (65) holds.

18 18 TSOGTGEREL GANTUMUR d) By definition, there is g : (a, b) R, continuous at x, with g(x) = f (x) 0, such that f(z) = f(x) + g(z)(z x) for z (a, b). (73) Since g is continuous at x, we infer the existence if an open interval (c, d) x such that g(z) 0 for all z (c, d). For t f((c, d)), we have z = f 1 (t) (c, d), and f 1 (t) f 1 (y) = z x = f(z) f(x) g(z) = t y g(f 1 (t)). (74) The function 1 g(f 1 (t)) is continuous at t = y, meaning that f 1 is differentiable at y, and that (66) holds. Exercise 4.7. Prove a) of the preceding theorem. Example 4.8. (a) By the product rule, we have (x 2 ) = 1 x + x 1 = 2x, (x 3 ) = (x 2 x) = 2x x + x 2 1 = 3x 2,... (x n ) = nx n 1 (n N). (75) (b) By the sum and product rules, all polynomials are differentiable in R, and the derivative of a polynomial is again a polynomial. (c) Given a function f : (a, b) R that does not vanish anywhere in (a, b), we can write the reciprocal function 1 f as F f with F (z) = 1 z. If f is differentiable at x (a, b), then by the chain rule, 1 f is differentiable at x and In particular, we have ( 1 ) (x) = (F f) (x) = F (f(x))f (x) = f (x) f [f(x)] 2. (76) (x n ) = nxn 1 x 2n = nx n 1 (n N). (77) (d) Let f(x) = x n for x [0, ), where n N. We have f (x) = nx n 1 at x > 0, and the inverse function is the arithmetic n-the root f 1 (y) = n y (y 0). Since f (x) > 0 for x > 0, the inverse f 1 is differentiable at each y > 0, with (f 1 ) (y) = Moreover, by the chain rule, for m Z and n N, we infer that is (x m n ) = (( n x) m ) = m( n x) m 1 1 n x 1 n n 1 f (f 1 (y)) = 1 n( n y) n 1 = 1 n y 1 n n. (78) = m n x m 1 n + 1 n n = m n x m n 1, (79) (x a ) = ax a 1 at each x > 0, for a Q. (80) Exercise 4.9. Let f, g : (a, b) R be functions differentiable at x (a, b), with g(x) 0. Show that the quotient f/g is differentiable at x, and the following quotient rule holds. Compute the derivative of q(x) = 3x3 x (f ) (x) f (x)g(x) f(x)g (x) = g [g(x)] 2. (81)

19 FUNCTIONS OF A REAL VARIABLE Applications of differentiation The following result is sometimes called the first derivative test. Lemma 5.1. Let f : (a, b) R be a function, and let c (a, b) be a maximum of f, in the sense that f(x) f(c) for all x (a, b). Suppose that f is differentiable at c. Then f (c) = 0. Proof. Suppose that f (c) 0. By definition of differentiability, there exists a function g : (a, b) R, continuous at c, with g(c) = f (c), such that f(x) = f(c) + g(x)(x c), x (a, b). (82) If f (c) > 0, we let x n = c + 1 n, and if f (c) < 0, we let x n = c 1 n, for n N. Then by continuity, for sufficiently large n, we have g(x n ) g(c) 1 2 g(c). This means that g(x n )(x n c) 1 2n f (c) or f(x) f(c) + 1 2n f (c), contradicting the maximality of f(c). Remark 5.2. At least in principle, the first derivative test gives a way to find the maximums and minimums of a differentiable function. Namely, let f C ([a, b]) be given. Then the extreme value theorem (Theorem 3.25) guarantees the existence of a maximum ξ [a, b]. If ξ (a, b) and if f is differentiable in (a, b), then f (ξ) = 0. In other words, all maximums located in the interior (a, b) can be found by comparing the values f(c) at the critical points, which are by definition the solutions c (a, b) of the equation f (c) = 0. The consideration of critical points leads to the following fundamental result, known as Rolle s theroem. It was proved by Michel Rolle in Theorem 5.3 (Rolle). Let f C ([a, b]) be differentiable at each x (a, b), with f(a) = f(b). Then there exists ξ (a, b) such that f (ξ) = 0. Proof. The extreme value theorem (Theorem 3.25) yields the existence of ξ [a, b] with the property that f(x) f(ξ) for all x [a, b]. Without loss of generality, we can assume that f is not constant, and that ξ (a, b), because if f(x) f(a) for all x [a, b], then we can replace f by f. Then the first derivative test (Lemma 5.1) implies that f (ξ) = 0. The following important consequence was proved by Lagrange in Theorem 5.4 (Mean value). Let f C ([a, b]) be differentiable at each x (a, b). Then there exists ξ (a, b) such that f(b) f(a) = f (ξ)(b a). Proof. Define the function F : [a, b] R by f(b) f(a) F (x) = f(x) (x a). (83) b a We have F (a) = F (b) = f(a), F C ([a, b]), and F is differentiable at each x (a, b), with F (x) = f f(b) f(a) (x). b a (84) By Rolle s theorem (Theorem 5.3), there exists ξ (a, b) such that F (ξ) = 0. Remark 5.5. If f is continuous at c, then for y close to c, we have f(y) = f(c) + e, with e 0 as y c 0. Thus we can use f(c) to approximate f(y), but there is a very little information on the size of the error e. If in addition, f is differentiable at c, then we have f(y) = f(c) + f (c)(y c) + e 1, (85) with e 1 vanishing faster than y c as y c 0. This shows that e = f (c)(y c) + e 1, but we still do not have a precise quantitative information on e 1. The mean value theorem (Theorem 5.4) reveals a quantitative bound on the error e, provided that f is differentiable in a region (not only at the point c), even when c and y are at a finite distance from each other.